2. A solution is 0.01 M in Ba2+ and 0.01 M in Ca2+. Sodium sulfate is added to selectively precipitate one of the cations, while leaving the other in solution. a) Given that BaSO4 and CaSO4 have Ksp values of 1.07 x 10-10 and 7.10 x 10-5 respectively, which cation will precipitate first? Explain why. b) What is the minimum concentration of SO42⁻ that will trigger the precipitation of the cation that precipitates first?

Explanation:

The solution contain  0.01 M concentration of Ba²⁺

0.01M concentration of Ca²⁺

Ksp ( solubility constant) for BaSO₄ = 1.07 × 10⁻¹⁰

Ksp for CaSO₄ = 7.10 × 10⁻⁵

(BaSO₄) = (Ba²⁺) (SO₄²⁻)

1.07 × 10⁻¹⁰ = 0.01 M (SO₄²⁻)

1.07 × 10⁻¹⁰ / 0.01 = ( SO₄²⁻)

1.07 × 10⁻⁸ M = ( SO₄²⁻)

so the minimum of concentration of concentration sulfate needed is 1.07 × 10⁻⁸ M

For CaSO₄

CaSO₄ = ( Ca²⁺) (  SO₄²⁻)

7.10 × 10⁻⁵  = 0.01 (SO₄²⁻)

(SO₄²⁻) = 7.10 × 10⁻⁵   / 0.01 = 7.10 × 10⁻³ M

so BaSO₄ will precipitate first since its cation (0.01 M Ba²⁺) required a less concentration of SO₄²⁻ (1.07 × 10⁻⁸ M ) compared to CaSO₄

b) The minimum concentration of SO₄²⁻ that will trigger the precipitation of the cation ( 0.01 M Ba²⁺) that precipitates first is 1.07 × 10⁻⁸ M

b. is the answer hoped i